Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());

        double[] p = new double[n];

        for (int i = 0; i < n; i++) {
            p[i] = Double.parseDouble(sc.next());
        }

        double[][] dp = new double[n + 1][n + 1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i][j] = 0;
            }
        }

        dp[0][0] = 1;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i + 1; j++) {
                dp[i + 1][j] += dp[i][j] * (1 - p[i]);
                dp[i + 1][j + 1] += dp[i][j] * p[i];
            }
        }

        double result = 0;
        for (int i = (n / 2) + 1; i <= n; i++) {
            result += dp[n][i];
        }

        System.out.println(result);
    }
}